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\topic{Lecture 11 \\Multiple Integral\\ \scriptsize Change of Order (07 Oct 2009)}
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\section{Change of Order}
As, we have seen that in evaluation of a double integral, either we integrate function with respect to $y$ first and then with respect to $x$, or we may integrate with respect to $x$ first and then with respect to $y$. For region $R$, in former case, we take limits of integration by taking a strip parallel to the $y$-axis, and in the letter case by taking one parallel to $x$-axis.

When it is required to change the order of integration in an integral for which the limits are given, we first decide the region $R$ of integration. Knowing the region of integration, we can then put in the limits for integration in the reverse order.
\begin{example}
Change the order of the integration in the integral 
\[\int_0^{a \cos \alpha} \int_{x \tan \alpha}^{\sqrt{(a^2-x^2)}} f(x,y) dxdy\]
\end{example}

The given limits show that the region of integration is bounded by the curves $y=\tan x$, $y=\sqrt{(a^2-x^2)}$, $x=0$ and $x=a \cos \alpha$.

The first curve $y=\tan x$ is a line through the origin and the second curve is $y=\sqrt{(a^2-x^2)}$, the circle. These produce the point of intersection at $(a \cos \alpha, a \sin \alpha)$. There fore the region of integration is $OPQ$ in figure*

When we integrate with respect to $x$ first along horizontal line, the strip starts from $x=0$. But some of strips end on $OP$ while the other end on $PQ$. Now the line of demarkation is $y=a \sin \alpha$ so the region must be divided into two sub regions 
\[x=0,~x=\cot \alpha,~y=0,~y=a \sin \alpha\]
and
\[x=0,~x=\sqrt{(a^2-y^2)},~y=a \sin \alpha,~y=a\]
Hence on changing the order of integration, the integral is equal to
\[\int_0^{a \sin \alpha} \int_{0}^{y \cot \alpha} f(x,y) dxdy + \int_{a \cos \alpha}^a \int_{0}^{\sqrt{(a^2-y^2)}} f(x,y) dxdy\]
\section*{Problems}
\begin{enumerate}
\item Evaluate $\int _{0}^{\infty}\int _{x}^{\infty }\frac{e^{-y} }{y} dxdy $
\item Change the order of the integration in the double integral 
\[\int _{0}^{2a}\int _{\sqrt{2ax-x^{2} }}^{\sqrt{2ax}} V dx dy \]
\item Changing the order of integration of $\int _{0}^{\infty } \int _{0}^{\infty } e^{-xy} \sin nx~dxdy$ show that \[\int _{0}^{\infty } \frac{\sin nx}{x} dx=\frac{\pi }{2}\]
\item By changing the order of integration, evaluate:
\[\int _{0}^{a} \int _{y^{2}/a}^{y} \frac{y}{(a-x)\sqrt{ax-y^{2} } } dxdy\]
\end{enumerate}

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